## Calculus 8th Edition

The given integral is \begin{aligned} \int_{-2}^{0}\left(x^{2}+x\right) d x &=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x \\ & =\frac{2}{3} \end{aligned}
$$\int_{-2}^{0}\left(x^{2}+x\right) d x$$ The width of the subintervals is $$\Delta x=(0-(-2)) / n=\frac{2}{n},$$ and $$x_{i}=-2+i \Delta x=-2+\frac{2i}{n}.$$ The given integral is \begin{aligned} \int_{-2}^{0}\left(x^{2}+x\right) d x &=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x \\ & =\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(-2+\frac{2 i}{n}\right) \frac{2}{n} \\ &=\lim _{n \rightarrow \infty} \frac{2}{n} \sum_{i=1}^{n}\left[\left(-2+\frac{2 i}{n}\right)^{2}+\left(-2+\frac{2 i}{n}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{2}{n} \sum_{i=1}^{n}\left[4-\frac{8 i}{n}+\frac{4 i^{2}}{n^{2}}-2+\frac{2 i}{n}\right] \\ &=\lim _{n \rightarrow \infty} \frac{2}{n} \sum_{i=1}^{n}\left(\frac{4 i^{2}}{n^{2}}-\frac{6 i}{n}+2\right) \\ & =\lim _{n \rightarrow \infty} \frac{2}{n}\left[\frac{4}{n^{2}} \sum_{i=1}^{n} i^{2}-\frac{6}{n} \sum_{i=1}^{n} i+\sum_{i=1}^{n} 2\right] \\ &=\lim _{n \rightarrow \infty}\left[\frac{8}{n^{3}} \frac{n(n+1)(2 n+1)}{6}-\frac{12}{n^{2}} \frac{n(n+1)}{2}+\frac{2}{n} \cdot n(2)\right] \\ & =\lim _{n \rightarrow \infty}\left[\frac{4}{3} \frac{(n+1)(2 n+1)}{n^{2}}-6 \frac{n+1}{n}+4\right]\\ &=\lim _{n \rightarrow \infty}\left[\frac{4}{3} \frac{n+1}{n} \frac{2 n+1}{n}-6\left(1+\frac{1}{n}\right)+4\right]\\ & =\lim _{n \rightarrow \infty}\left[\frac{4}{3}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)-6\left(1+\frac{1}{n}\right)+4\right]\\ &=\frac{4}{3}(1)(2)-6(1)+4 \\ & =\frac{2}{3} \end{aligned}